Differentiability and Differentiation MCQ Questions & Answers in Calculus | Maths

$$f\left( x \right) = \left| {x - \pi } \right|\left( {{e^{\left| x \right|}} - 1} \right)\sin \left| x \right|$$
Check differentiability of $$f\left( x \right)$$  at $${x = \pi }$$  and $$x=0$$
at $$x = \pi :$$
$$\eqalign{ & {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {x + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0}}{h} = 0 \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {x - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0}}{{ - h}} = 0 \cr & \because {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} = {\text{L}}{\text{.H}}{\text{.D}}{\text{.}} \cr} $$
Therefore, function is differentiable at $$x = \pi $$
at $$x=0 :$$
$$\eqalign{ & {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - 0}}{h} = 0 \cr & {\text{L}}{\text{.H}}{\text{.D}}{\text{.}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| { - h} \right| - 0}}{{ - h}} = 0 \cr & \therefore {\text{R}}{\text{.H}}{\text{.D}}{\text{.}} = {\text{L}}{\text{.H}}{\text{.D}}{\text{.}} \cr} $$
Therefore, function is differentiable.
at $$x=0.$$
Since, the function $$f\left( x \right)$$  is differentiable at all the points including $$z$$ and 0. i.e., $$f\left( x \right)$$  is every where differentiable .
Therefore, there is no element in the set $$S.$$
$$ \Rightarrow S = \phi \,\,\left( {{\text{an empty set}}} \right)$$

As per question,
$$p=$$  left hand derivation of $$\left| {x - 1} \right|$$   at $$x = 1\,\, \Rightarrow p = - 1$$
Also $$\mathop {\lim }\limits_{x \to 1^+ } g\left( x \right) = p$$
Where $$g\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log \,{{\cos }^m}\left( {x - 1} \right)}},\,0 < x < 2,\,m,\,n$$         are integers, $$m \ne 0,\,n > 0$$
$$\therefore $$ We get,
$$\eqalign{ & \mathop {\lim }\limits_{x \to {1^ + }} \frac{{{{\left( {x - 1} \right)}^n}}}{{\log \,{{\cos }^m}\left( {x - 1} \right)}} = - 1 \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{h^n}}}{{\log \,{{\cos }^m}h}} = - 1 \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{{h^n}}}{{m\left( {\log \,\cos h} \right)}} = - 1\,\,\left[ {{\text{Using L'Hospital's rule}}} \right] \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{n\,{h^{n - 1}}\cos \,h}}{{m\left( { - \sin \,h} \right)}} = - 1\,\,\left[ {{\text{Using L'Hospital's rule}}} \right] \cr & \Rightarrow \mathop {\lim }\limits_{h \to 0} \frac{{n\,{h^{n - 2}}\cos \,h}}{{m\left( {\frac{{\sin \,h}}{h}} \right)}} = 1 \cr & \Rightarrow n = 2\,\,{\text{and}}\,\,\,m = 2 \cr} $$

$$\eqalign{ & f\left( x \right) = - {p^3}{x^3} - 6{p^3}.\sin \,x + \left( {6{p^2} + p} \right)\cos \,x \cr & \therefore f'\left( x \right) = - 3{p^3}{x^2} - 6{p^3}.\cos \,x - \left( {6{p^2} + p} \right)\sin \,x \cr & \therefore f''\left( x \right) = - 6{p^3}x + 6{p^3}.\sin \,x - \left( {6{p^2} + p} \right)\cos \,x \cr & \therefore f'''\left( x \right) = - 6{p^3} + 6{p^3}.\cos \,x + \left( {6{p^2} + p} \right)\sin \,x \cr & \therefore f'''\left( 0 \right) = - 6{p^3} + 6{p^3}.1 = 0 = {\text{independent of}}\,{\text{ }}p. \cr} $$

$$\eqalign{ & s = \sqrt {{t^2} + 1} \cr & \Rightarrow \frac{{ds}}{{dt}} = \frac{t}{{\sqrt {{t^2} + 1} }} \cr & \Rightarrow \frac{{{d^2}s}}{{d{t^2}}} = \frac{1}{{\sqrt {{{\left( {\sqrt {{t^2} + 1} } \right)}^3}} }} \cr & \Rightarrow \frac{{{d^2}s}}{{d{t^2}}} = \frac{1}{{{s^3}}} \cr} $$

$$\eqalign{ & 5f\left( x \right) + 3f\left( {\frac{1}{x}} \right) = x + 2 \cr & \Rightarrow 5f\left( {\frac{1}{x}} \right) + 3f\left( x \right) = \frac{1}{x} + 2 \cr & {\text{From these, 16}}f\left( x \right) = 5x - \frac{3}{x} + 4 \cr & \therefore 16f'\left( x \right) = 5 + \frac{3}{{{x^2}}} \cr & \therefore y = xf\left( x \right) \Rightarrow \frac{{dy}}{{dx}} = f\left( x \right) + xf'\left( x \right) \cr & = \frac{1}{{16}}\left( {5x - \frac{3}{x} + 4} \right) + x.\frac{1}{{16}}\left( {5 + \frac{3}{{{x^2}}}} \right) \cr & \therefore \,\,{\text{at }}x = 1,\,\frac{{dy}}{{dx}} = \frac{1}{{16}}\left( {5 - 3 + 4} \right) + \frac{1}{{16}}\left( {5 + 3} \right) = \frac{7}{8} \cr} $$

$$\eqalign{ & {\text{Given relation is }}{\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2},\,c > 0 \cr & {\text{let }}x - a = c\,\cos \,\theta {\text{ and }}y - b = c\,\sin \,\theta \cr & {\text{Therefore,}} \cr & \frac{{dx}}{{d\theta }} = - c\,\sin \,\theta {\text{ and }}\frac{{dy}}{{d\theta }} = c\,\cos \,\theta \cr & \therefore \,\frac{{dy}}{{dx}} = - \cot \,\theta \cr & {\text{Differentiable both sides with respect to}}\,\theta {\text{, we get}} \cr & \frac{d}{{d\theta }}\left( {\frac{{dy}}{{dx}}} \right) = \frac{d}{{d\theta }}\left( { - \cot \,\theta } \right) \cr & {\text{or, }}\frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right)\frac{{dx}}{{d\theta }} = {\text{cose}}{{\text{c}}^2}\theta \cr & {\text{or, }}\frac{{{d^2}y}}{{d{x^2}}}\left( { - c\,\sin \,\theta } \right) = {\text{cose}}{{\text{c}}^2}\theta \cr & {\text{or, }}\frac{{{d^2}y}}{{d{x^2}}} = - \frac{{{\text{cose}}{{\text{c}}^2}\theta }}{c} \cr & \therefore \,\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}} = \frac{{c{{\left[ {1 + {{\cot }^2}\theta } \right]}^{\frac{3}{2}}}}}{{ - {\text{cose}}{{\text{c}}^3}\theta }} = \frac{{c{{\left( {{\text{cose}}{{\text{c}}^2}\theta } \right)}^{\frac{3}{2}}}}}{{ - {\text{cose}}{{\text{c}}^3}\theta }} = - c \cr & {\text{which is constant and is independent of }}a{\text{ and }}b. \cr} $$

$$\left| {x - a} \right|$$  is not differentiable at $$x = a.$$
Also $$\tan \,x$$  is not differentiable if $$x = \left( {2k + 1} \right)\frac{\pi }{2},\,k\, \in \,I$$
$$\therefore $$  In the interval $$\left( {0,\,2} \right),\,f\left( x \right)$$   is not derivable at $$x = 0.5,\,x = 1$$    and $$x = \frac{\pi }{2}.$$

$$\eqalign{ & {\text{We have }}f'\left( {{0^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( 0 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\left| {\cos \frac{\pi }{h}} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} h\left| {\cos \frac{\pi }{h}} \right| \cr & = 0 \times {\text{ some finite value}} = 0 \cr & {\text{Also }}f'\left( {{0^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 - h} \right) - f\left( 0 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}\left| {\cos \frac{\pi }{{ - h}}} \right|}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} - h\left| {\cos \frac{\pi }{h}} \right| \cr & = 0 \times {\text{ some finite value}} = 0 \cr & \because f'\left( {{0^ + }} \right) = f'\left( {{0^ - }} \right) \cr & \Rightarrow f\,{\text{is differentiable at }}x = 0 \cr & {\text{Now }}f'\left( {{2^ + }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2 + h} \right) - f\left( 2 \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}\left| {\cos \frac{\pi }{{2 + h}}} \right| - 4\left| {\cos \frac{\pi }{2}} \right|}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}\left( {\cos \frac{\pi }{{2 + h}}} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h}\sin \left( {\frac{\pi }{2} - \frac{\pi }{{2 + h}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h}\sin \left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right) \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 + h} \right)}^2}}}{h} \times \frac{{\sin \left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right)}}{{\left( {\frac{{\pi h}}{{2\left( {2 + h} \right)}}} \right)}} \times \frac{{\pi h}}{{2\left( {2 + h} \right)}} = \pi \cr & {\text{Also }}f'\left( {{2^ - }} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {2 - h} \right) - f\left( 2 \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}\left| {\cos \frac{\pi }{{2 - h}}} \right| - 0}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{ - {{\left( {2 - h} \right)}^2}\cos \left( {\frac{\pi }{{2 - h}}} \right)}}{{ - h}} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}\sin \left( {\frac{\pi }{2} - \frac{\pi }{{2 - h}}} \right)}}{h} \cr & = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {2 - h} \right)}^2}}}{h} \times \frac{{\sin \left( {\frac{{ - \pi h}}{{2\left( {2 - h} \right)}}} \right)}}{{\left( {\frac{{ - \pi h}}{{2\left( {2 - h} \right)}}} \right)}} \times \frac{{ - \pi h}}{{2\left( {2 - h} \right)}} = - \pi \cr} $$
As $$f'\left( {{2^ + }} \right) \ne f'\left( {{2^ - }} \right)\,\, \Rightarrow f$$     is not differentiable at $$x=2.$$

Since. $$\left[ {{x^3}} \right]$$ is not continuous and differentiable at integral points. So, $$f\left( x \right)$$  is continuous and differentiable in $$\left[ {4,\,6} \right]$$  if $$\left[ {\frac{{{{\left( {x - 2} \right)}^3}}}{a}} \right] = 0\,\, \Rightarrow a \geqslant 64$$

$$\eqalign{ & f\left\{ {g\left( x \right)} \right\} \Rightarrow \frac{{df\left\{ {g\left( x \right)} \right\}}}{{dg\left( x \right)}}.g'\left( x \right) = 1 \cr & \Rightarrow \sin \left\{ {g\left( x \right)} \right\}.g'\left( x \right) = 1 \cr & \Rightarrow g'\left( x \right) = \frac{1}{{\sin \left\{ {g\left( x \right)} \right\}}} \cr} $$